Solve z+2=sqrt{6z+19}

$$z + 2 = \sqrt { 6 z + 19 }$$

Solve for z

$$z=5$$

Solution Steps

Square both sides of the equation.

$$\left(z+2\right)^{2}=\left(\sqrt{6z+19}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(z+2\right)^{2}.

$$z^{2}+4z+4=\left(\sqrt{6z+19}\right)^{2}$$

Calculate \sqrt{6z+19} to the power of 2 and get 6z+19.

$$z^{2}+4z+4=6z+19$$

Subtract 6z from both sides.

$$z^{2}+4z+4-6z=19$$

Combine 4z and -6z to get -2z.

$$z^{2}-2z+4=19$$

Subtract 19 from both sides.

$$z^{2}-2z+4-19=0$$

Subtract 19 from 4 to get -15.

$$z^{2}-2z-15=0$$

To solve the equation, factor z^{2}-2z-15 using formula z^{2}+\left(a+b\right)z+ab=\left(z+a\right)\left(z+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite factored expression \left(z+a\right)\left(z+b\right) using the obtained values.

$$\left(z-5\right)\left(z+3\right)$$

To find equation solutions, solve z-5=0 and z+3=0.

$$z=5 z=-3$$

Substitute 5 for z in the equation z+2=\sqrt{6z+19}.

$$5+2=\sqrt{6\times 5+19}$$

Simplify. The value z=5 satisfies the equation.

$$7=7$$

Substitute -3 for z in the equation z+2=\sqrt{6z+19}.

$$-3+2=\sqrt{6\left(-3\right)+19}$$

Simplify. The value z=-3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Equation z+2=\sqrt{6z+19} has a unique solution.

$$z=5$$